Timeline for Identity map between $L^2(\mu)$ and $L^2(\mu_{\rm sf})$
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Apr 1, 2018 at 16:13 | comment | added | p4sch | You can just take $A=E$! And any other measurable $A \subset E$ satisfies $\mu(A) \leq \mu(E)$. | |
| Apr 1, 2018 at 10:07 | comment | added | Student | @p4sch Why if $\mu(E) < \infty$, we have $\mu_{sf}(E) = \mu(E)$? Thank you for your explanation | |
| Mar 24, 2018 at 16:57 | history | edited | p4sch | CC BY-SA 3.0 |
Corrected the intial mistake!
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| Mar 24, 2018 at 16:01 | comment | added | YCor | (There a typographic problem with the use of "rm" in the formula. In makes undesired roman letters.) | |
| Mar 24, 2018 at 15:57 | comment | added | YCor | @NateEldredge thanks: deleted my fake counterexample (one interest of looking for the "simplest" counterexample is that when it fails, it's easier to see!) | |
| Mar 24, 2018 at 15:46 | comment | added | p4sch | Nate, you are right! I have ignored that we only talk about equivalence classes of functions. I will correct my answer ... | |
| Mar 24, 2018 at 14:19 | comment | added | Nate Eldredge | @YCor: Likewise, in your example of an infinite singleton, $L^2(\mu_{\rm sf})$ is zero-dimensional, not one-dimensional. It does contain all the constants, but they are all a.e. equal to zero. So although a function like $1$ is not in $L^2(\mu)$, we can choose a better representative of its $\mu_{\rm sf}$ equivalence class which is in $L^2(\mu)$ (here, the zero function). If the conjecture is true, the question would be how to make the "right" choice more generally. | |
| Mar 24, 2018 at 13:59 | comment | added | Nate Eldredge | So surjectivity is the following statement: for any $g \in L^2(\mu_{\rm sf})$ there exists $f \in L^2(\mu)$ such that $f=g$, $\mu_{\rm sf}$-a.e. | |
| Mar 24, 2018 at 13:53 | comment | added | Nate Eldredge | I don't agree with the counterexample. Since $\mu_{\rm sf} = 0$, we have $1_A = 0$, $\mu_{\rm sf}$-a.e. Indeed, $L^2(\mu_{\rm sf}) = 0$, i.e., its only element is a single equivalence class containing all the $\Sigma$-measurable functions on $\Omega$, and this class certainly is in the image of the map, because the (class of the) zero function in $L^2(\mu)$ maps to it. | |
| Mar 24, 2018 at 10:49 | history | edited | p4sch | CC BY-SA 3.0 |
Added some comments.
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| Mar 24, 2018 at 10:39 | comment | added | Amir Sagiv | So, the conclusion is that even for positive measures it is not necessarily bijective? I'm a bit confused... | |
| Mar 24, 2018 at 10:31 | history | edited | p4sch | CC BY-SA 3.0 |
Additonal Comment!
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| Mar 24, 2018 at 10:29 | review | First posts | |||
| Mar 24, 2018 at 10:40 | |||||
| Mar 24, 2018 at 10:26 | history | answered | p4sch | CC BY-SA 3.0 |